Answer:
The 99.9% confidence interval for the population proportion is (0.548, 0.896).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 72, \pi = \frac{x}{n} = \frac{52}{72} = 0.722[/tex]
99.9% confidence level
So [tex]\alpha = 0.001[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.001}{2} = 0.9995[/tex], so [tex]Z = 3.29[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.722 - 3.29\sqrt{\frac{0.722*0.278}{72}} = 0.548[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.722 + 3.29\sqrt{\frac{0.722*0.278}{72}} = 0.896[/tex]
The 99.9% confidence interval for the population proportion is (0.548, 0.896).
